Calculate the magnitude of ring strain energy in (kJ/mol) of cyclopropane from the for the following  data :

${\mathrm{\Delta }}_{f}H\left[C_3{H}_6\left(g\right)\right]=55;{\mathrm{\Delta }}_{f}H\left[\mathrm{C}\right(g\left)\right]=715.0$

$\begin{array}{r}{\mathrm{\Delta }}_{f}H\left[H\right(g\left)\right]=220;Be(C-C)=355;BE(C-H)=410\\ \text{ (all in }\mathrm{kJ}/\text{ mole) }\end{array}$

Given that

$3\mathrm{C}\left(\mathrm{s}\right)+3\mathrm{C}\left(\mathrm{g}\right)\mathrm{\Delta }{H}_2=3\times 715$   $=2145\mathrm{kJ}/\mathrm{mol}$

$3{\mathrm{H}}_2\left(\mathrm{s}\right)\to 6\mathrm{H}\left(\mathrm{g}\right)$  $\mathrm{\Delta }{H}_3=6\times 220 = 1320 \mathrm{kJ}/{\mathrm{mol}}^{-1}$

$3\mathrm{C}\left(g\right)+4\mathrm{H}\left(g\right)\to$ $∆{\mathrm{H}}_4$

$\mathrm{\Delta }{H}_4=-\mathrm{B}\cdot \mathrm{E}\cdot (3\mathrm{C}-\mathrm{C}+6\mathrm{C}-\mathrm{H})+$$\text{ Ring Strain Energy (RSE) }$

$=\mathrm{\Delta }{H}_1-\mathrm{\Delta }{H}_2-\mathrm{\Delta }{H}_3$

$\begin{array}{r}\text{ R.S.E. }=\mathrm{\Delta }{H}_1-\mathrm{\Delta }{H}_2-\mathrm{\Delta }{H}_3+\text{ B.E. }(3\mathrm{C}-\mathrm{C}+6\mathrm{C}-\mathrm{H})\\ =55-2145-1320+(3\times 355+6\times 410)\\ =115\mathrm{kJ}/\mathrm{mol}\end{array}$